# Bash multiplication and addition

```
for k in {0..49};
do
a=$(($((2*$k))+1));
echo $a;
done
```

Hi, I need a simplified expression for the third line, maybe one that does not use command substitution.

Best Answer

Using arithmetic expansion.

```
for (( k = 0; k < 50; ++k )); do
a=$(( 2*k + 1 ))
echo "$a"
done
```

Using the antiquated `expr`

utility.

```
for (( k = 0; k < 50; ++k )); do
a=$( expr 2 '*' "$k" + 1 )
echo "$a"
done
```

Using `bc -l`

( `-l`

not actually needed in this case as no math functions are used).

```
for (( k = 0; k < 50; ++k )); do
a=$( bc -l <<<"2*$k + 1" )
echo "$a"
done
```

Using `bc -l`

as a co-process (it acts like a sort of computation service in the background¹).

```
coproc bc -l
for (( k = 0; k < 50; ++k )); do
printf "2*%d + 1\n" "$k" >&${COPROC[1]}
read -u "${COPROC[0]}" a
echo "$a"
done
kill "$COPROC_PID"
```

That last one looks (arguably) cleaner in `ksh93`

.

```
bc -l |&
bc_pid="$!"
for (( k = 0; k < 50; ++k )); do
print -p "2*$k + 1"
read -p a
print "$a"
done
kill "$bc_pid"
```

¹ This solved a an issue for me once where I needed to process a large amount of input in a loop. The processing required some floating point computations, but spawning `bc`

a few times in the loop proved to be exceedingly slow. Yes, I could have solved it in many other ways, but I was bored...